Mathematics

Mathematics

Problem

Four real constants a,b,A,Ba, b, A, B are given and f(θ)=1−acos⁡θ−bsin⁡θ−Acos⁡2θ−Bsin⁡2θ.f(\theta) = 1 - a \cos \theta - b \sin \theta - A \cos 2\theta - B \sin 2\theta. Prove that if f(θ)≥0,∀θ∈Rf(\theta) \ge 0, \forall \theta \in R, then a2+b2≤2a^2 + b^2 \le 2 and A2+B2≤1A^2 + B^2 \le 1.
Intuitive explanation

The key is to recognize that f(θ)≥0f(\theta) \geq 0 for all θ\theta means the trigonometric expression has a global minimum of at least zero. By strategically choosing specific values of θ\theta—particularly those that isolate or amplify certain terms—you can extract constraints on the coefficients. The double-angle terms cos⁡2θ\cos 2\theta and sin⁡2θ\sin 2\theta can be rewritten using cos⁡2θ\cos^2\theta and sin⁡2θ\sin^2\theta, creating a connection between first and second-order terms that forces bounds on both (a,b)(a,b) and (A,B)(A,B).

Intuitive explanation

The key is to recognize that f(θ)≥0f(\theta) \geq 0 for all θ\theta means the trigonometric expression has a global minimum of at least zero. By strategically choosing specific values of θ\theta—particularly those that isolate or amplify certain terms—you can extract constraints on the coefficients. The double-angle terms cos⁡2θ\cos 2\theta and sin⁡2θ\sin 2\theta can be rewritten using cos⁡2θ\cos^2\theta and sin⁡2θ\sin^2\theta, creating a connection between first and second-order terms that forces bounds on both (a,b)(a,b) and (A,B)(A,B).

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